I'm assuming that you meant to write 1,1,2-triphenyl-2-bromo-1-propanol.

You are correct that water will leave and generate a very stable benzylic cation which is further stabilized by the adjacent bromine. This will form a three membered ring, bromonium ion. The positive charge will reside on both the 1 and 2 carbons. This will probably not revert back to the vinyl compound through loss of Br+ (Br+ is a poor leaving group) even if Br- is present. What may happen is that the water might re-add to the bromonium ion system and regenerate the original bromohydrin. This will undoubtedly occur by the E1 mechanism and the fleeting, isolated charge (and thus the hydroxyl) will reside on the most stable carbonium ion... on the diphenyl-substituted carbon (the "1" position), regenerating starting material.

You might consider that the H+ adds to the bromine (in the neutral compound) and then loses HBr leaving behind a stabilized cation alpha to a hydroxyl group. If this happens, the oxygen of the adjacent alcohol could migrate over to form an epoxide and regenerate a proton. What do we know about epoxides and strong acids? Strong acids add to epoxides in a Markovnikov fashion and the product would be 112-triphenyl-1-bromo-2-propanol, a new product. This is unlikely to happen however since the hydroxyl is much more basic (a likely target for H+) than bromine. Even if it did happen, the first reaction (adding H+ to OH group) would definitely occur and regenerate 1,1,2-tripheny-2-bromo-1-propanol.

Therefore, if I had to provide the answer, I would submit the bromonium ion intermediate as the only reasonable answer.